Overall heat transfer. An aluminum soda can is pulled from the refrigerator at 3

Question

Overall heat transfer. An aluminum soda can is pulled from the refrigerator at 35 F and placed in a Styrofoam can holder. The can (and holder) is placed in a room at 70 F. The Styrofoam is 0.5" thick, and has a conductivity of 0.033 W/m-C. The heat transfer coefficient inside the can (from the soda to the aluminum wall) is 50 W/m^2-C, while the heat transfer coefficient from the holder to the air is 8 W/m^2-C. What is the heat loss per area through the side of the can? -37 W/m^2 +14 W/m^2 -19 W/m^2 +120 W/m^2 +68 W/m^2

Explanation / Answer

Aluminium can temperature = 35 deg.F, F=(9/5)*C+32, C= (35-32)*5/9= 1.7 deg.c

Room temperature= 70deg.F, C= (70-32)*5/9= 21.11 deg.c

Driving force for heat transfer= 21.11-1.70 = 19.41 deg.c

There are three resistances in series.

They are Styrofoam, two convective heat transfer coefficients

Overall resistance 1/h= sum of individual resistances = thickness/thermal conductivity+1/50+1/8

Thickness=0.5inches= 0.5/12 ft (12 inches= 1ft)= 0.042ft = 0.042*0.3048 m ( 1ft=0.3048) = 0.013m

Hence overall resistance.1/U= 0.013/0.033+0.02+0.125 =0.54

U= 1/0.54= 1.85 W/m2.deg.c

Heat transfer rate= U*driving force = 1.85*19.41= 36 W/m2 ( so close answer is A)

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